Minimize swaps to rearrange array such that parity of index and corresponding element is same

• Last Updated : 09 Apr, 2021

Given an array A[], the task to find minimum swapping operations required to modify the given array A[] such that for every index in the array, parity(i) = parity(A[i]) where parity(x) = x % 2. If it’s impossible to obtain such an arrangement, then print -1.

Examples:

Input: A[] = { 2, 4, 3, 1, 5, 6 }
Output:
Explanation:
Swapping (4, 3) and (5, 6) modifies the array to [2, 3, 4, 1, 6, 5] such that the parity of i and A[i] is same for all indices.

Input: A[] = {1, 2, 5, 7}
Output: -1
Explanation:
The given array cannot be rearranged as per required condition.

Approach:
To solve the problem mentioned above an optimal approach is to choose such an index where parity(i) and parity(A[i]) aren’t the same.

• Initialize two variables needodd and needeven to 0 which will store the parity of each element. Check the parity of index if it’s odd then increase needodd value by 1 otherwise increase needeven.
• If needodd and needeven aren’t same, then required arrangement is not possible.
• Otherwise, the final result is obtained by needodd variable, as it is the number of operations that are required. This is because, at any moment, we choose an odd element whose parity is not the same with parity of their index and similarly choose an even element and swap them.

Below is the implementation of the above approach:

C++

 // C++ implementation to minimize// swaps required to rearrange// array such that parity of index and// corresponding element is same#include using namespace std; // Function to return the// parity of numberint parity(int x){    return x % 2;     } // Function to return minimum// number of operations requiredint solve(int a[], int size){         // Initialize needodd and    // needeven value by 0    int needeven = 0;    int needodd = 0;         for(int i = 0; i < size; i++)    {        if(parity(i) != parity(a[i]))        {                         // Check if parity(i) is odd            if(parity(i) % 2)            {                                 // increase needodd                // as we need odd no                // at that position.                needodd++;            }            else            {                 // increase needeven                // as we need even                // number at that position                needeven++;            }        }    }         // If needeven and needodd are unequal    if(needeven != needodd)        return -1;    else        return needodd;} // Driver Codeint main(){    int a[] = { 2, 4, 3, 1, 5, 6};    int n = sizeof(a) / sizeof(a);     // Function call    cout << solve(a, n) << endl;     return 0;} // This code is contributed by venky07

Java

 // Java implementation to minimize // swaps required to rearrange // array such that parity of index and // corresponding element is sameimport java.util.*; class GFG{     // Function to return the// parity of numberstatic int parity(int x){    return x % 2;} // Function to return minimum// number of operations requiredstatic int solve(int a[], int size){         // Initialize needodd and    // needeven value by 0    int needeven = 0;    int needodd = 0;         for(int i = 0; i < size; i++)    {        if(parity(i) != parity(a[i]))        {                         // Check if parity(i) is odd            if(parity(i) % 2 == 1)            {                                 // Increase needodd                // as we need odd no                // at that position.                needodd++;            }            else            {                                 // Increase needeven                // as we need even                // number at that position                needeven++;            }        }    }         // If needeven and needodd are unequal    if(needeven != needodd)        return -1;    else        return needodd;} // Driver Codepublic static void main (String[] args){    int a[] = { 2, 4, 3, 1, 5, 6};    int n = a.length;         // Function call    System.out.println(solve(a, n));}} // This code is contributed by offbeat

Python3

 # Python3 implementation to minimize# swaps required to rearrange# array such that parity of index and# corresponding element is same # Function to return the# parity of numberdef parity(x):    return x % 2 # Function to return minimum# number of operations required def solve(a, size):         # Initialize needodd and    # needeven value by 0    needeven = 0    needodd = 0    for i in range(size):                         if parity(i)!= parity(a[i]):                         # Check if parity(i) is odd            if parity(i) % 2:                                 # increase needodd                # as we need odd no                # at that position.                needodd+= 1            else:                # increase needeven                # as we need even                # number at that position                needeven+= 1         # If needeven and needodd are unequal    if needodd != needeven:        return -1             return needodd     # Driver code    if __name__ =="__main__":         a = [2, 4, 3, 1, 5, 6]    n = len(a)    print(solve(a, n))

C#

 // C# implementation to minimize // swaps required to rearrange // array such that parity of index and // corresponding element is sameusing System;class GFG{     // Function to return the// parity of numberstatic int parity(int x){  return x % 2;} // Function to return minimum// number of operations requiredstatic int solve(int[] a, int size){      // Initialize needodd and  // needeven value by 0  int needeven = 0;  int needodd = 0;   for(int i = 0; i < size; i++)  {    if(parity(i) != parity(a[i]))    {      // Check if parity(i) is odd      if(parity(i) % 2 == 1)      {        // Increase needodd        // as we need odd no        // at that position.        needodd++;      }      else      {        // Increase needeven        // as we need even        // number at that position        needeven++;      }    }  }   // If needeven and needodd are unequal  if(needeven != needodd)    return -1;  else    return needodd;} // Driver Codepublic static void Main (){  int[] a = {2, 4, 3, 1, 5, 6};  int n = a.Length;   // Function call  Console.Write(solve(a, n));}} // This code is contributed by Chitranayal

Javascript


Output:
2

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